Indefinite Integrals Trig Quiz - Powers, Fractions & Trig

1) What is the antiderivative of f(x) = 0?

0

1

C (any constant)

X

The derivative of any constant function is 0. Therefore, the antiderivative of 0 is any constant. We write ∫ 0 dx = C, where C is an arbitrary constant. Option A, 0, is one specific antiderivative, but the general antiderivative includes all constants.

Explanation

The derivative of any constant function is 0. Therefore, the antiderivative of 0 is any constant. We write ∫ 0 dx = C, where C is an arbitrary constant. Option A, 0, is one specific antiderivative, but the general antiderivative includes all constants.

2) Evaluate: ∫ (x⁷ - 4x³ + x) dx

(1/8)x⁸- x⁴ + (½)x² + C

7x⁶ - 12x² + 1 + C

(1/8)x⁸- (1/4)x⁴ + (½)x² + C

(1/8)x⁸- x⁴ + x² + C

Apply the power rule to each term. For x⁷: ∫x⁷ dx = (1/8)x^8. For -4x³: ∫-4x³ dx = -4*(1/4)x⁴ = -x⁴. For x: ∫x dx = (½)x². Combine and add C: (1/8)x⁸- x⁴ + (½)x² + C.

Explanation

Apply the power rule to each term. For x⁷: ∫x⁷ dx = (1/8)x^8. For -4x³: ∫-4x³ dx = -4*(1/4)x⁴ = -x⁴. For x: ∫x dx = (½)x². Combine and add C: (1/8)x⁸- x⁴ + (½)x² + C.

3) Which operation is the inverse of differentiation?

Integration

Multiplication

Composition

Logarithm

Differentiation finds the derivative of a function. Integration, specifically finding the indefinite integral, finds a function whose derivative is given. This reverses the process of differentiation. Therefore, integration is the inverse operation of differentiation.

Explanation

Differentiation finds the derivative of a function. Integration, specifically finding the indefinite integral, finds a function whose derivative is given. This reverses the process of differentiation. Therefore, integration is the inverse operation of differentiation.

4) Evaluate: ∫ (½) eˣ dx

(½) eˣ + C

2eˣ + C

(½) e^(x+1) + C

Eˣ + C

The integral of eˣ is eˣ. For a constant multiple, we have ∫ k * f(x) dx = k * ∫ f(x) dx. Here, k = 1/2. So ∫ (½) eˣ dx = (½) ∫ eˣ dx = (½) eˣ + C.

Explanation

The integral of eˣ is eˣ. For a constant multiple, we have ∫ k * f(x) dx = k * ∫ f(x) dx. Here, k = 1/2. So ∫ (½) eˣ dx = (½) ∫ eˣ dx = (½) eˣ + C.

5) Find ∫ (5x⁴ + 2/x²) dx.

X⁵ - 2/x + C

20x³ - 4/x³ + C

X⁵ + 2/x + C

X⁵ - 2/x³ + C

Rewrite the integrand: 5x⁴ + 2/x² = 5x⁴ + 2x^(-2). Now integrate term by term. ∫5x⁴ dx = 5*(1/5)x⁵ = x⁵. ∫2x^(-2) dx = 2 * [x^(-2+1)/(-2+1)] = 2 * [x^(-1)/(-1)] = -2x^(-1) = -2/x. Add C: x⁵ - 2/x + C.

Explanation

Rewrite the integrand: 5x⁴ + 2/x² = 5x⁴ + 2x^(-2). Now integrate term by term. ∫5x⁴ dx = 5*(1/5)x⁵ = x⁵. ∫2x^(-2) dx = 2 * [x^(-2+1)/(-2+1)] = 2 * [x^(-1)/(-1)] = -2x^(-1) = -2/x. Add C: x⁵ - 2/x + C.

6) The indefinite integral ∫ f(x) dx represents:

The area under the curve of f(x) from a to b.

The slope of the tangent line to f(x).

The general antiderivative of f(x).

The derivative of f(x).

By definition, the indefinite integral ∫ f(x) dx is the collection of all antiderivatives of f(x). It is expressed as F(x) + C, where F'(x) = f(x). Option A describes a definite integral. Option B describes the derivative. Option D is the operation opposite to integration.

Explanation

By definition, the indefinite integral ∫ f(x) dx is the collection of all antiderivatives of f(x). It is expressed as F(x) + C, where F'(x) = f(x). Option A describes a definite integral. Option B describes the derivative. Option D is the operation opposite to integration.

7) Evaluate: ∫ (3 sin(x) - 4 cos(x)) dx

-3 cos(x) - 4 sin(x) + C

3 cos(x) + 4 sin(x) + C

-3 cos(x) + 4 sin(x) + C

3 cos(x) - 4 sin(x) + C

We know ∫ sin(x) dx = -cos(x) and ∫ cos(x) dx = sin(x). So, ∫3 sin(x) dx = 3*(-cos(x)) = -3 cos(x). ∫ -4 cos(x) dx = -4 * sin(x) = -4 sin(x). Combine: -3 cos(x) - 4 sin(x) + C.

Explanation

We know ∫ sin(x) dx = -cos(x) and ∫ cos(x) dx = sin(x). So, ∫3 sin(x) dx = 3*(-cos(x)) = -3 cos(x). ∫ -4 cos(x) dx = -4 * sin(x) = -4 sin(x). Combine: -3 cos(x) - 4 sin(x) + C.

8) If F'(x) = f(x), then f(x) is called the ____________ of F(x).

Derivative

Antiderivative

Integral

Constant

This is the definition of derivative. If F'(x) = f(x), then f(x) is the derivative of F(x). The antiderivative of f(x) would be F(x). The integral of f(x) would be F(x) + C.

Explanation

This is the definition of derivative. If F'(x) = f(x), then f(x) is the derivative of F(x). The antiderivative of f(x) would be F(x). The integral of f(x) would be F(x) + C.

9) Evaluate: ∫ ( (x² + 1) / x ) dx, for x > 0.

(½)x² + ln|x| + C

X + 1/x + C

(½)x² + 1/x² + C

X²/2 + ln x + C

First simplify the integrand: (x² + 1)/x = x²/x + 1/x = x + 1/x. Now integrate term by term: ∫ x dx = (½)x². ∫ (1/x) dx = ln|x|. Since x > 0, this is ln(x). So the result is (½)x² + ln(x) + C.

Explanation

First simplify the integrand: (x² + 1)/x = x²/x + 1/x = x + 1/x. Now integrate term by term: ∫ x dx = (½)x². ∫ (1/x) dx = ln|x|. Since x > 0, this is ln(x). So the result is (½)x² + ln(x) + C.

10) Which of the following functions does NOT have an elementary antiderivative?

X⁵

Sin(3x)

E^(-x²)

1/(x²+1)

While x⁵, sin(3x), and 1/(x²+1) have antiderivatives expressible in terms of elementary functions (polynomials, trigonometric, and arctangent respectively), the function e^(-x²) is known to not have an antiderivative that can be written as a finite combination of elementary functions. Its integral is related to the error function, which is not elementary.

Explanation

While x⁵, sin(3x), and 1/(x²+1) have antiderivatives expressible in terms of elementary functions (polynomials, trigonometric, and arctangent respectively), the function e^(-x²) is known to not have an antiderivative that can be written as a finite combination of elementary functions. Its integral is related to the error function, which is not elementary.

11) Evaluate: ∫ 6 sec²(x) dx

6 tan(x) + C

6 sec(x) tan(x) + C

3 tan²(x) + C

6 cos²(x) + C

We recall that the derivative of tan(x) is sec²(x). Therefore, the antiderivative of sec²(x) is tan(x). For a constant multiple, ∫ 6 sec²(x) dx = 6 ∫ sec²(x) dx = 6 tan(x) + C.

Explanation

We recall that the derivative of tan(x) is sec²(x). Therefore, the antiderivative of sec²(x) is tan(x). For a constant multiple, ∫ 6 sec²(x) dx = 6 ∫ sec²(x) dx = 6 tan(x) + C.

12) The power rule for integration, ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, is valid for:

All real numbers n.

All real numbers n except n = -1.

Only positive integers n.

Only rational numbers n.

The power rule works for any real number n ≠ -1. When n = -1, the integrand is 1/x, and the antiderivative is ln|x|, not a power function. The formula would involve division by zero if n = -1. So the exception is n = -1.

Explanation

The power rule works for any real number n ≠ -1. When n = -1, the integrand is 1/x, and the antiderivative is ln|x|, not a power function. The formula would involve division by zero if n = -1. So the exception is n = -1.

13) Evaluate: ∫ (5 - 2x + 3x²) dx

5x - x² + x³ + C

5x - 2x² + 3x³ + C

5x - x² + 3x³ + C

5x - 2x² + x³ + C

Integrate term by term. ∫5 dx = 5x. ∫ -2x dx = -2*(½)x² = -x². ∫ 3x² dx = 3*(⅓)x³ = x³. So the result is 5x - x² + x³ + C.

Explanation

Integrate term by term. ∫5 dx = 5x. ∫ -2x dx = -2*(½)x² = -x². ∫ 3x² dx = 3*(⅓)x³ = x³. So the result is 5x - x² + x³ + C.

14) Which statement provides the most precise definition of F(x) being an antiderivative of f(x)?

F(x) is the area under the curve of f(x).

F'(x) = f(x).

F'(x) = F(x).

F(x) = f(x) + C.

The definition of an antiderivative is based on reversing the process of differentiation. A function F(x) is called an antiderivative of f(x) if the derivative of F(x) is equal to f(x). Option A describes a definite integral, not an antiderivative. Option C reverses the relationship. Option D incorrectly relates the functions through addition rather than differentiation.

Explanation

The definition of an antiderivative is based on reversing the process of differentiation. A function F(x) is called an antiderivative of f(x) if the derivative of F(x) is equal to f(x). Option A describes a definite integral, not an antiderivative. Option C reverses the relationship. Option D incorrectly relates the functions through addition rather than differentiation.